Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{4r^2 - 20r}{r^2 + 7r} \times \dfrac{4r^2 + 4r - 168}{r^2 - 16r + 60} $
First factor out any common factors. $x = \dfrac{4r(r - 5)}{r(r + 7)} \times \dfrac{4(r^2 + r - 42)}{r^2 - 16r + 60} $ Then factor the quadratic expressions. $x = \dfrac {4r(r - 5)} {r(r + 7)} \times \dfrac {4(r - 6)(r + 7)} {(r - 6)(r - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {4r(r - 5) \times 4(r - 6)(r + 7) } {r(r + 7) \times (r - 6)(r - 10) } $ $x = \dfrac {16r(r - 6)(r + 7)(r - 5)} {r(r - 6)(r - 10)(r + 7)} $ Notice that $(r - 6)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {16r\cancel{(r - 6)}(r + 7)(r - 5)} {r\cancel{(r - 6)}(r - 10)(r + 7)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $x = \dfrac {16r\cancel{(r - 6)}\cancel{(r + 7)}(r - 5)} {r\cancel{(r - 6)}(r - 10)\cancel{(r + 7)}} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $x = \dfrac {16r(r - 5)} {r(r - 10)} $ $ x = \dfrac{16(r - 5)}{r - 10}; r \neq 6; r \neq -7 $